1/1.3+1/3.5+.....+1/2015.2017
K
Khách
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H
9 tháng 5 2016
B=1/1.3+1/3.5+...+1/2015.2017
B= 1/2 . 2. ( 1/1.3+1/3.5 + .... + 1/2015 .2017)
B = 1/2 . ( 2/1.3 + 2/3.5 + ......+ 2/2015.2017)
B = 1/2. ( 1/1+ -1/3 + 1/3 + -1/5 + 1/5 +....+ 1/2015 + -1/2017)
B= 1/2 . ( 1/1 + -1 / 2017) = 1/2 . 2016 / 2017 = 2016 / 4034
Vậy B = 2016 / 4034 nha bn.pham hong thai
9 tháng 8 2015
Dễ thôi:
Khoảng cách là 2
\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\frac{1}{2}.\left(1-\frac{1}{2017}\right)=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
KC
0
NK
13
LH
21 tháng 8 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
21 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
mk đầu tiên đấy
TT
0
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
27 tháng 2 2017
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25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\) \(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\) \(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\) \(=\dfrac{2016}{2017}\) |
Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2016}{2017}\)
\(=\frac{1008}{2017}\)